YEAR 12 Trial Exam Paper FURTHER MATHEMATICS. Written examination 1. Worked solutions

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1 YEAR 12 Trial Exam Paper 2016 FURTHER MATHEMATICS Written examination 1 s This book presents: worked solutions, giving you a series of points to show you how to work through the questions mark allocations tips on how to approach the exam This trial examination produced by Insight Publications is NOT an official VCAA paper for the 2016 Further Mathematics written examination 1. The Publishers assume no legal liability for the opinions, ideas or statements contained in this trial exam. This examination paper is licensed to be printed, photocopied or placed on the school intranet and used only within the confines of the purchasing school for examining their students. No trial examination or part thereof may be issued or passed on to any other party including other schools, practising or non-practising teachers, tutors, parents, websites or publishing agencies without the written consent of Insight Publications. Copyright Insight Publications 2016

2 SECTION A Core Data analysis Question 1 Answer: C The data is recorded as a numerical variable. Marks are allocated on a scale from 0 to 40, with scores given to the nearest whole mark. 2 Tips Numerical discrete data is represented by numbers that are counted. Numerical continuous data is represented by numbers that are measured rather than counted. Question 2 Answer: D A. Histogram a histogram does not allow us to consider numerical variables that have been split into categories. B. Segmented bar chart a segmented bar chart allows us to compare two categorical variables. C. Scatterplot a scatterplot is best used for considering the relationship between two numerical variables. D. Parallel boxplots parallel boxplots will allow us to display numerical data for two categories, and compare the distribution for these categories. E. Dot plot a single dot plot does not accommodate two categorical variables. Copyright Insight Publications 2016 SECTION A

3 Question 3 Answer: A In order to include an outlier in our description of this data distribution, we need to calculate the lower fence. Lower fence: Q IQR As the test score 3 is less than the lower fence, it can be considered an outlier. 3 Tip Often it is easier to use your calculator to get a statistical summary, then calculate the IQR and upper/lower fence. Copyright Insight Publications 2016 SECTION A

4 4 Question 4 Answer: D A. Option A is incorrect because, using quartiles to summarise numerical data, 75% of the data is below the third quartile. For this distribution, Q 3 is equal to 38. B. Option B is incorrect because the mean value for the distribution is Rounded to the nearest whole number, this is 30 C. Option C is incorrect because, for a negatively skewed distribution, the mean is less than the median. For this distribution, the mean is 30 and the median is 34 (to the nearest whole number). D. Option D is correct because it is not appropriate to use the median as a measure of centre when a distribution is skewed or contains an outlier. The distribution for Further Mathematics test scores is both skewed and contains an outlier. E. Option E is incorrect because it would also be appropriate to display the data for Further Mathematics test scores as a dot plot. Question 5 Answer: E The data is approximately normally distributed, we are therefore considering a bell-shaped distribution. A diagram is very useful for calculating our answer. In this question we are asked to consider students between 82 and 94. We first use the bell curve to calculate the percentage. Percentage between 82 and 94 = = 81.5% We now calculate 81.5% of 284 students. Number of students = Rounded to the nearest whole number, this is 231 students. Copyright Insight Publications 2016 SECTION A

5 Question 6 Answer: A A. Option A is correct because the range for A is 37, and the range for B is 33. *Hint: The range includes the outlier. B. Option B is incorrect because the boxplot for 12A is negatively skewed. C. Option C is incorrect because the median for 12A is higher than that for 12B. D. Option D is incorrect because: Q3 for 12A is it a 38, Q3 for 12B is at 32. Therefore, the top 25% of scores for 12A are higher than for 12B. E. Option E is incorrect because the outlier at 3 is the minimum value for 12A, this is lower than the minimum for 12B. 5 Question 7 Answer: B In this case, the explanatory variable is Further Mathematics test score. The response variable is Science test score. With the information given, we can use the following formulae to calculate the slope, b, and y-intercept, a, for our regression equation: r sy b sx a y bx b a a 9.97 b 0.73 where, r is the correlation coefficient s y is the standard deviation for y. s x is the standard deviation for x. x is the mean for x. y is the mean for y. Therefore, our regression equation is: Science Further Mathematics. Copyright Insight Publications 2016 SECTION A

6 Question 8 Answer: D The value of Pearson s correlation coefficient is a positive value, therefore suggesting a positive relationship. Its value is , therefore the relationship is strong. 6 Tips With an r value this high, a non-linear relationship would be obvious in the graph. For example, it may appear to be a curve, with points close together, as shown below: This graph has a high Pearson s correlation coefficient, similar to that in our question. Due to this high value, the strong non-linear relationship is obvious. Copyright Insight Publications 2016 SECTION A

7 Question 9 Answer: B Ensure that you enter Test 1 as your explanatory variable, x. Choose 2 points on the graph: For example, (2, 12) and (30, 32) Find the slope using rise run y y 2 1. x 2 x 1 Slope = Use the general formula for a linear equation: y y1 m( x x1) Equation is: y (x 2) y x 1.42 y 0.71x The equation is: y = x We need to make sure that we write this using our correct variables, however. Test Test 1 Copyright Insight Publications 2016 SECTION A

8 8 Question 10 Answer: B A. Option A is incorrect because the slope for the regression equation is positive; therefore, Test 2 scores will increase by the slope as Test 1 scores increase by one. B. Option B is correct because the value of the coefficient of determination is This is therefore a correct interpretation of our r 2 value. C. Option C is incorrect because we have no information to suggest this outcome. As we do not have a data value below 3 for Test 1, this is extrapolating. Also, we would look at the y-intercept to obtain information here. D. Option D is incorrect because we can use a regression analysis to look at correlation between two variables only. We need to be very careful that we do not use the word cause. E. Option E is incorrect. Because the data appears to form a linear relationship, there is no need to perform a data transformation. Question 11 Answer: E By matching points on our original scatterplot with those predicted by the linear regression line, we can eliminate options, as the residual values do not match. Question 12 Answer: C By matching points to the circle of transformations, we can see that the most appropriate options are those listed in option C. Copyright Insight Publications 2016 SECTION A

9 9 Question 13 Answer: C Question 14 Answer: C As there are 11 months of data recorded, the seasonal indices must add to 11. We therefore subtract our known seasonal indices from 11: Question 15 Answer: B Substitute the predicted sales figure for March 2016 into the equation: Multiply by the seasonal index for March, 0.76, to get the actual sales for March: Round to the nearest hundred: 5300 Copyright Insight Publications 2016 SECTION A

10 Question 16 Answer: C We wish to reduce the seasonal index for June from 1.5 to 1.0. Therefore we need to reduce the index by = % 1.5 We therefore wish to reduce the index by 33%. 10 Recursion and financial modelling Question 17 Answer: C We can calculate iterations by hand, using the recurrence relation. V0 7 V 2V V V 10 1 V V V V V V V V V V V V You can use your calculator to quickly define a recurrence relation, and find iterations. Tips Remember that for a recurrence relation, you must use the previous term to find your next term. For example, we use V 2 to find V 3. Do not keep using V 0 to find each term. Copyright Insight Publications 2016 SECTION A

11 11 Question 18 Answer: D Linear growth or decay will be represented by a straight line graph. This means we could use either option D or E. As we are modelling linear decay, the graph should be displayed as having a negative slope; option D is our answer. Options A and B are both curved, therefore representations of geometric growth and decay. Option C is an example of payments made at irregular intervals, and the payment is changing. This is also clearly a growth, rather than decay. Question 19 Answer: D The general model for representing flat rate depreciation is: r. 100 V 0 = initial value of the asset, V n 1 V n D, where D V0 In this scenario, we have the following: V , r 20 D Therefore, our recurrence relation is: V , V n 1 V n 6400 This is option D. Question 20 Answer: C Using our recurrence relation (we will show iterations on the calculator) Copyright Insight Publications 2016 SECTION A

12 Question 21 Answer: B We start by defining the terms in the rule for reducing-balance depreciation. 12 V Vn 18000, r R n 10 r Vn 1 V r n We can solve this on our calculator: Tips The calculator has given us two possible answers. A percentage yearly rate of 186% would reduce our value by a large amount over the first year: This is already below the value given to us for 10 years of use. Therefore, we must discount this answer as being impossible. Question 22 Answer: D The interest is calculated from the balance of the loan after the last payment. 1.5 Therefore: Interest = $ Note: In the initial information given, the interest rate is provided as an annual interest rate. However, we are calculating interest on a monthly basis. We must therefore convert our annual interest rate to a monthly rate 18% %. Copyright Insight Publications 2016 SECTION A

13 Question 23 Answer: E The interest is calculated from the balance of the loan after the last payment. 1.5 Therefore: Interest = $ The final payment must cover the balance owing and the interest. Therefore: Final payment = $ Question 24 Answer: C Start by calculating the future value for the investment after 6 months, at the interest rate of 6.1%. Remember to change your PpY and CpY to 12. Next, we will change our present value to the value of the investment after 6 months, and calculate the future value after another 6 months. We must change our interest rate to 5.75 Remember to change this value to a negative, as you are investing this money in the bank. Finally, calculate interest earned by subtracting the initial value, , from the future value after 12 months. Copyright Insight Publications 2016 SECTION A

14 SECTION B Modules Module 1 Matrices Question 1 Answer: B Incorrect elements in each matrix are highlighted below. A. B. 14 C. D. E. Question 2 Answer: A A triangular matrix will have zero elements either above or below the main diagonal. In this case, we have zero elements above the main diagonal; this is a lower triangular matrix. Copyright Insight Publications 2016 SECTION B Module 1

15 Question 3 Answer: E We are looking for a matrix multiplication that will result in a 3 1 matrix. 15 A. B C D E This matrix multiplication is undefined: The columns in the first matrix do not equal the rows in the second. This matrix multiplication is undefined: The columns in the first matrix do not equal the rows in the second. This matrix multiplication is undefined: The columns in the first matrix do not equal the rows in the second This matrix multiplication is undefined: The columns in the first matrix do not equal the rows in the second. This matrix multiplication is defined: The columns in the first matrix equal the rows in the second. This matrix multiplication will result in the following: giving a total for stores A, B and C. Copyright Insight Publications 2016 SECTION B Module 1

16 Question 4 Answer: B To solve the simultaneous equations: AX Step 1: Find the inverse, A 1. Use your calculator for this. B 16 Step 2: Multiply matrix B by the inverse of matrix A. Copyright Insight Publications 2016 SECTION B Module 1

17 17 Question 5 Answer: D To find S 3, we use the transition matrix, T. As our initial matrix is S 1, this is information for our first week. Therefore, we only require 2 transitions to find the third week. S 3 T 2 S 1 On the calculator: Tips Your numbers here must add up to the initial number, 215. This will require you to think carefully about your rounding. Beware of questions where the initial state matrix is S 1. Your transitions n1 will need to be one less than the state matrix you require: S T S. n 1 Copyright Insight Publications 2016 SECTION B Module 1

18 Question 6 Answer: D We are looking for two consecutive matrices between which there is no change in state. This can be seen below for S 30 and S Question 7 Answer: C For a recurrence relation of this form, we must work out S 2 first, then use this in the equation to find S 3. Step 1: Find S 2. S2 S1T B Step 2: Find S. 3 S3 S2T B Copyright Insight Publications 2016 SECTION B Module 1

19 19 Question 8 Answer: C The two-step dominance matrix can be found by squaring the one-step dominance matrix. One-step dominance is found when a student has a direct link to another, with the arrow facing towards their opponent. For example, student A defeated student B. Therefore a 1 is placed in the matrix. The one-step dominance matrix is shown below: A B C D E A B C D E We can square this to find the two-step dominances, as shown on the calculator page below. Copyright Insight Publications 2016 SECTION B Module 1

20 Module 2 Networks and decision mathematics Question 1 Answer: E 20 Question 2 Answer: D For a network to contain a Euler trail, the network must contain exactly 2 vertices with odd degree, with the rest of the vertices having an even degree. Network A has 2 vertices with even degree, and the rest are odd this does not satisfy the conditions for an Eulerian trail. Network B has 2 vertices with odd degree, and the rest are even this satisfies the conditions for an Eulerian trail. Network C: has 5 vertices with odd degree, and only one even this does not satisfy conditions for Euler trail. Copyright Insight Publications 2016 SECTION B Module 2

21 21 Question 3 Answer: A The minimum spanning tree is shown below: The length of this spanning tree is = 84 Question 4 Answer: A The shortest path can be found using one of two methods: Method 1: By inspection Copyright Insight Publications 2016 SECTION B Module 2

22 22 Method 2: Using Dijkstra s algorithm A B C D E F Exit Ent = 120 = 175 = 120 = D C = 570 E = 435 F Explanatory notes 1. Create a table with the starting vertex (entrance) as the first row vertex. 2. Write the other vertices as column vertices. 3. Write down the distance from entrance to all other vertices. 4. Put a box around the smallest number in the row. 5. Look at the column vertex for this box number; this becomes the next row vertex. 6. Continue process until the destination is reached. 7. Backtrack from the smallest number in row F, to the destination. Copyright Insight Publications 2016 SECTION B Module 2

23 23 Question 5 Answer: D Maximum flow = minimum cut = = 15 Copyright Insight Publications 2016 SECTION B Module 2

24 Question 6 Answer: E Use the Hungarian algorithm. Step 1: row reduction Subtract the smallest number from each row, from every number in the same row. 24 Mathematics task to complete Student Statistics Financial Networks Geometry maths Andrea Bryan Carmen David Step 2: column reduction Subtract the smallest number from any column that does not already contain a 0 from each number in the column. Mathematics task to complete Student Statistics Financial Networks Geometry maths Andrea Bryan Carmen David Copyright Insight Publications 2016 SECTION B Module 2

25 25 Step 3: If the minimum number of lines required to cover all the zeros in the table is equal to the number of allocation to be made, draw a directed bipartite graph, with an edge for every zero value in the table Step 4: Use the bipartite graph to make allocations. Andrea only has one possible task Statistics. David now only has Geometry remaining. Carmen only had one possible task Financial maths. Therefore, Bryan will do the remaining task Networks. Copyright Insight Publications 2016 SECTION B Module 2

26 26 Question 7 Answer: D Question 8 Answer: B The critical path is made up of those activities that have no float/slack time. These are our critical activities, which must be completed on time, in order for the project to be completed in the minimum time. Copyright Insight Publications 2016 SECTION B Module 2

27 27 Module 3 Geometry and measurement Question 1 Answer: C As the sails are similar triangles, we can calculate the scale ratio for the triangles The area of the smaller triangle is equal to 1.65 m 2. We can multiply this by square of the scale ratio in order to calculate the area of the larger triangle. 2 2 Area larger triangle = m 2 We must remember to square the scale ratio as we are calculating area, which is in units squared. Question 2 Answer: B We start by labelling the sides of the triangle, in relation to the angle we are trying to find: m We have been given the numerical value of the opposite side and the adjacent side, and we therefore use tan opp adj Tips Your calculator settings must be in Degrees for this calculation. If your calculator is in Radians, you have 2 possible options: Change your settings to degrees Convert your answer to degrees, using the appropriate function in your CAS calculator. Copyright Insight Publications 2016 SECTION B Module 3

28 Question 3 Answer: C Step 1: Calculate the length of the base of the triangle created by the rod. Find the length of the diagonal within the rectangular base: 28 Pythagoras theorem: c a b We are trying to find c: c a b Step 2: Find the length of the rod, by creating a vertical triangle within the box. We are trying to find c: c a 2 b cm Copyright Insight Publications 2016 SECTION B Module 3

29 Question 4 Answer: D Step 1: Calculate the height of the trapezium, using Pythagoras theorem. 29 The base of the triangle is: m m We are trying to find b: b c a Step 2: Calculate the TSA by calculating area of individual shapes. The trapezium 1 area abh 2 1 area area We have 2 trapeziums: area The rectangular sides area length width area 125 area 60 Again, we have 2 of these: area The rectangular base area length width area 412 area 48 Total surface area Copyright Insight Publications 2016 SECTION B Module 3

30 30 Question 5 Answer: B Step 1: Calculate the volume of the cake. V l w h V V 2160 cm 3 Step 2: Calculate one quarter of this volume Step 3: Find the value of d, using the formula for the volume of a rectangular prism. V lw h Question 6 Answer: B The two cities have the same longitude, therefore they are on the same great circle. The total angle of both points representing the cities is (38 23) 61 We now find the arc length between these two points. The formula for arc length = s r 180 When looking at the Earth as our sphere, the radius is 6400 km. The angle,, is equal to 61. Therefore, the length of the arc s km 6814 km Copyright Insight Publications 2016 SECTION B Module 3

31 31 Question 7 Answer: E The diagram shows the plane of the meridian 13 E. N W E S First, we calculate the angle from the point at Rome to the South Pole. (4290) 132 The length of the arc s r km This distance from Rome to the South Pole along the great circle is km. Question 8 Answer: A The side length OM, is equal to the radius of the hemisphere. It is therefore 6 m. The length of OA is the difference between the radius of the hemisphere, 6, and the height of the water, 2.5: OA m m m To find the length of side AM, we use Pythagoras theorem: b c a To find the area of the surface of the water, we use A r 2 : Area = m 2 Copyright Insight Publications 2016 SECTION B Module 3

32 Module 4 Graphs and relations Question 1 Answer: C We start by selecting two coordinates on our graph: (0, 4 and 6, 8) Use these points to find the gradient of the straight line: m rise run y y x 2 x We can see that the y-intercept for this straight line is 4, from the graph. 2 Therefore, our equation in gradient intercept form is : y x 4 3 The equations given in the multiple choice options are in intercept form. Write the equation using a common denominator: 2x 12 y 3 Multiply both sides by 3: 3y 2x 12 Subtract 2x from both sides: 3y 2x 12 Hence, we have option C. 32 Question 2 Answer: A A. (1, 3) 4 (1) 2 (3) True C. (3, 7) (7) False E. (5, 6) (6) False B. (2, 9) (9) False D. (4, 5) (5) False Copyright Insight Publications 2016 SECTION B Module 4

33 Question 3 Answer: D A 35 minute appointment will fall into the third interval, which sits at $40, for 25 t Tips Pay attention to the open and closed circles in the step graphs. An open circle corresponds to a greater than or less than, but not equal to. A closed circle corresponds to a greater than, less than or equal to. Question 4 Answer: C We first need to find the value of k, in the equation. We can use the point (80, 64) to solve for k. Now, we can use the equation braking distance (speed)2 to find the braking distance when the speed is equal to Braking distance = m Question 5 Answer: B Average speed = total distance travelled total time Average speed = km/h 9 Copyright Insight Publications 2016 SECTION B Module 4

34 34 Question 6 Answer: C This can be done in one of two ways: algebraically and graphically. Algebraically: We can solve for the value of n, where our second equation equals the Cost equation. Note: This requires you to check where the first equation ends within the Revenue function, and realise that the Cost function will intersect the second equation after n = 150. Graphically: We can graph the three functions, and find the intersection between the Cost function, and the Revenue function. Question 7 Answer: B This is best done by looking at a graph of the linear equations specified by the programming problem. The feasible region can be seen below, as shown by the shaded region. Copyright Insight Publications 2016 SECTION B Module 4

35 35 Question 8 Answer: D A. Option A is incorrect because both the revenue and the cost equation have their highest power of x as 1. They are equations for a straight line graph, and therefore linear. B. Option B is incorrect because the revenue equation tells us that we will sell each copy for $75. C. Option C is incorrect because: C x D. Option D is correct because: Profit = Revenue Cost P 75x x P 63x P P E. Option E is incorrect because the revenue is equal to the cost when 190 copies are sold (using rounding to the nearest textbook), and after that, the revenue is more. We can see this by graphing the Revenue and Cost functions. END OF WORKED SOLUTIONS Copyright Insight Publications 2016